This is my answer to the same question posed on the Mathematics Stack Exchange. It is therefore licenced under CC-BY-SA 3.0.

Question

When entered into Wolfram|Alpha, \(\infty^{\infty}\) results in “complex infinity”. Why?

Answer

WA’s ComplexInfinity is the same as Mathematica’s: it represents a complex “number” which has infinite magnitude but unknown or nonexistent phase. One can use DirectedInfinity to specify the phase of an infinite quantity, if it approaches infinity in a certain direction. The standard Infinity is the special case of phase 0. Note that Infinity is different from Indeterminate (which would be the output of e.g. 0/0).

Some elucidating examples:

• 0/0 returns Indeterminate, since (for instance) the limit may be approached as \(\frac{1/n}{1/n}\) or \(\frac{2/n}{2/n}\), resulting in two different real numbers.
• 1/0 returns ComplexInfinity, since (for instance) the limit may be approached as \(\frac{1}{-1/n}\) or as \(\frac{1}{1/n}\), but every possible way of approaching the limit gives an infinite answer.
• Abs[1/0] returns Infinity, since the limit is guaranteed to be infinite and approached along the real line in the positive direction.

In your particular example, you get ComplexInfinity because the infinite limit may be approached as (e.g.) \(n^n\) or as \(n^{n+i}\).